Introduction

In this blog post, we will explore how to solve the LeetCode problem "805" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array nums. You should move each element of nums into one of the two arrays A and B such that A and B are non-empty, and average(A) == average(B). Return true if it is possible to achieve that and false otherwise. Note that for an array arr, average(arr) is the sum of all the elements of arr over the length of arr. Example 1: Input: nums = [1,2,3,4,5,6,7,8] Output: true Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5. Example 2: Input: nums = [3,1] Output: false Constraints: 1 <= nums.length <= 30 0 <= nums[i] <= 104

Explanation

Here's a breakdown of the approach and the Python code:

• Key Idea: The core idea is to check if there exists a subset of nums whose average is equal to the average of the entire array. If such a subset exists, the remaining elements will also have the same average, fulfilling the problem's condition.

• Transformation: To avoid floating-point comparisons and potential precision issues, we transform the problem. Let sum_nums be the sum of all elements in nums, and n be the length of nums. We need to find a subset A of size k (where 1 <= k < n) such that sum(A) / k == sum_nums / n. This is equivalent to sum(A) * n == sum_nums * k.

• Subset Sum Variation: This transformed condition turns into a variation of the subset sum problem. We iterate through possible subset sizes k from 1 to n - 1, and for each k, we check if there exists a subset of size k whose sum equals (sum_nums * k) / n. We can efficiently solve this using dynamic programming.

• Complexity:

  • Runtime: O(n² * max(nums)), where n is the length of nums and max(nums) is the maximum value in nums.
  • Storage: O(n * max(nums))

Code

    def splitArraySameAverage(nums):
    n = len(nums)
    sum_nums = sum(nums)

    for k in range(1, n):
        if (sum_nums * k) % n == 0:
            target_sum = (sum_nums * k) // n
            if possible_subset_sum(nums, k, target_sum):
                return True

    return False

def possible_subset_sum(nums, k, target_sum):
    dp = {}

    def solve(index, count, current_sum):
        if (index, count, current_sum) in dp:
            return dp[(index, count, current_sum)]

        if count == 0:
            return current_sum == 0

        if index >= len(nums):
            return False

        # Option 1: Exclude nums[index]
        exclude = solve(index + 1, count, current_sum)

        # Option 2: Include nums[index]
        include = solve(index + 1, count - 1, current_sum - nums[index])

        dp[(index, count, current_sum)] = exclude or include
        return dp[(index, count, current_sum)]

    return solve(0, k, target_sum)